One Vector Field

As has been explained in the previous section we are first going to solve for the one vector field case, not considering a Modulus field in the action. It turns out that it is usefull to introduce a additional coupling constant $a$, that governs the coupling strength between the dilaton and the vector field. For example, $a=0$ corresponds to the Einstein-Maxwell action and $a=1$ to the string effective action without the Modulus. The equations of motion can be solved for general $a$ (that's off course why it is usefull to introduce this extra parameter) and so we can compare all kind of different models with eachother (different values of $a$ can be obtained from all kind of different models).

So the action we are explicitly going to solve for in this one vector field case is (when $a=1$ it should be noticed that when we eventually want to use $SL(2,R)$ transformations, Axion terms should be added)

$$S=\int d^4 x \sqrt{-g} [-R +{\textstyle{1\over 2}}(\partial \phi)^2 - {\textstyle{1\over 4}}e^{a \phi} F^2]$$ (5.10)

We mentioned allready that for the $a=1$ case, the Axion constraint tells us we only have to consider singly charged solutions, but what about other values of $a$? It turns out that for values of $a$ different from one, doubly charged solutions (from now on called dyonic) don't exist. This follows from the equations of motion and asymptotic flatness which constrains the metric function $R(r)$ to be of a specific form. We'll show this to you explicitly afterwards, but first we are going to solve for singly charged (let's say electric) solutions. The equations to solve are

$$( \lambda R R')' -1 = -{\textstyle{1\over 4}}e^{a \phi} F_{rt}^{2} R^2 \, ,$$ (5.11)

$$\frac{-4 R''}{R} = ( \tau' )^2 \, ,$$ (5.12)

$$e^{a \phi} R^2 F_{rt} = c \, ,$$ (5.13)

$$(\lambda R^2 \phi')' = -{\textstyle{1\over 2}}a e^{a \phi} F_{rt}^2 R^2 \, .$$ (5.14)

There are now two ways to proceed, the first one is by demanding the electric field to be a Coulomb field and thus making an assumption for $A_t$ (i.e. $A_t \sim \frac{1}{r}$; remember $F_{rt}==A'_t=\partial _r A_t$ ). The second one is to take a closer look at the metric function $R(r)$. We know that eq.5.5 still holds and we have demanded the metric to be assymptotically flat, which means that $R^2(r) \approx r^2$ and $\lambda (r) \approx 1$ for large $r$. Let's assume the metric has to be of the following general form:

$$R^2(r) = ( r + \Sigma )^2 \left( \frac{r - \Sigma}{r+\Sigma} \right)^{2b} \quad \lambda (r) = \frac{ (r-r_+)(r-r_-)}{R^2}.$$ (5.15)

The equations of motion can now be integrated and the solution is the same Garfinkle, Horowitz and Strominger [23] found:

$$e^{\phi} (r) = e^{\phi_0} \left( \frac{r+\Sigma}{r-\Sigma} \right)^{\frac{-2a}{1+a^2}} \, ,$$

$$F_{rt} = \frac{Q}{(r+\Sigma)^2} \quad\quad\quad\mbox{with}\quad Q=c e^{-a \phi_0}\, ,$$ (5.16)

$$R^2 (r) = ( r + \Sigma )^2 \left( \frac{r - \Sigma}{r+\Sigma} \right)^\frac{2a^2}{1+a^2} \, .$$

From eq.(5.12) it follows that

$$\Sigma = r_- \quad \lor \quad \Sigma = r_+$$

and if we now make a translation in $r$: $r \rightarrow r-r_-$ and redefine our constants:

$$r_+' = r_+ + r_- \quad\mbox{and}\quad r_-' = 2 r_- \, ,$$

we can write the solution in the more familiar form (dropping the primes):

$$A_t = - \frac{Q}{r} \, ,$$ (5.17)

$$e^\phi = e^{\phi_0} \left( 1 - \frac{r_-}{r} \right)^{\frac{-2a}{1+a^2}} \, ,$$ (5.18)

$$R^2 (r) = r^2 \left( 1 -\frac{r_-}{r} \right)^\frac{2a^2}{1+a^2} \, ,$$ (5.19)

$$\lambda (r) = \left( 1 - \frac{r_+}{r} \right) \left( 1 - \frac{r_-}{r} \right)^{\frac{1-a^2}{1+a^2}} \, .$$ (5.20)

with the conditions:

$$Q^2 = e^{-a \phi_0} \frac{4}{1+a^2} r_- r_+ \, ,$$

$$2M = r_+ + \frac{1 - a^2}{1+a^2} r_- \, .$$

where $M$ can be interpreted as the mass in the assymptotic weak field limit, see the Black Holes section.

Note that we could have replaced the electric charge with a magnetic charge and the solution would be the same except that the dilaton would have changed sign ( $\phi \rightarrow -\phi$ ). This discrete transformation also is a subgroup of $SL(2,R)$ transformations and so this subgroup is a symmetry for general $a$.